Question

Two blocks of equal masses are placed on a horizontal surface. The surface of $A$ is smooth but that of $B$ has a friction coefficient of $0.2$ with the floor. Block $A$ is given a speed $5\text{m/s}$, towards $B$ which is kept at rest. Find the distance travelled by $B$, if the collision is perfectly elastic.
(take $g=10{\text{m/s}}^2$ )

• A. $\frac{5}{4}\text{m}$
• B. $5\text{m}$
• C. $\frac{25}{4}\text{m}$
• D. $\frac{5}{2}\text{m}$

Answer 1

The correct option is C $\frac{25}{4}\text{m}$


Given
Initial speed of Block $A,u=5\text{m/s}$
After collision:



Assume
$v_1=$ Speed of Block $A$ towards right
$v_2=$ Speed of Block $B$ towards right
Just after the collision speed of Block $B$ is
By momentum conservation along the motion of the blocks (because no external force is present along that direction)
$\Rightarrow mu+0=m{v}_1+m{v}_2$
$\Rightarrow u=v_1+v_2$ ...(1)
Also we know that,
$e=\frac{\text{Velocity of seperation}}{\text{Velocity of approach}}$
$\Rightarrow e=\frac{v_2-v_1}{u-0}=1$ (As because of elastic collision)
$\Rightarrow v_2-v_1=u$ ...(2)
From eq (1) and (2) we have
$v_2=u,v_1=0$
Again, a friction force is acting on Block $B$, so, the situation can be shown below


From equation of motion we have
$v^2-u^2=2as$
Here, final speed $v=0$ and $u=5\text{m/s}$
also, $a=-\frac{\mu mg}{m}=-\mu g=-0.2\times 10=-2{\text{m/s}}^2$
Putting the values in the above equation we have
$0-5^2=2\times (-2)\times s$
$\Rightarrow s=\frac{25}{4}\text{m}.$
Hence, the distance travelled by $B$ is $\frac{25}{4}\text{m}.$